Problem: Simplify and expand the following expression: $ \dfrac{4n + 1}{n - 3}-\dfrac{n}{2n - 10} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(n - 3)(2n - 10)$ Multiply the first term by $\dfrac{2n - 10}{2n - 10}$ $ \begin{align*} \dfrac{4n + 1}{n - 3} \times \dfrac{2n - 10}{2n - 10} & = \dfrac{(4n + 1)(2n - 10)}{(n - 3)(2n - 10)} \\ & = \dfrac{8n^2 - 38n - 10}{(n - 3)(2n - 10)}\end{align*} $ Multiply the second term by $\dfrac{n - 3}{n - 3}$ $ \begin{align*} \dfrac{n}{2n - 10} \times \dfrac{n - 3}{n - 3} & = \dfrac{(n)(n - 3)}{(2n - 10)(n - 3)} \\ & = \dfrac{n^2 - 3n}{(2n - 10)(n - 3)}\end{align*} $ Now we have: $ = \dfrac{8n^2 - 38n - 10}{(n - 3)(2n - 10)} - \dfrac{n^2 - 3n}{(2n - 10)(n - 3)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{8n^2 - 38n - 10 - (n^2 - 3n)}{(n - 3)(2n - 10)} $ $ = \dfrac{8n^2 - 38n - 10 - n^2 + 3n}{(n - 3)(2n - 10)} $ $ = \dfrac{7n^2 - 35n - 10}{(n - 3)(2n - 10)}$ Expand the denominator: $ = \dfrac{7n^2 - 35n - 10}{2n^2 - 16n + 30}$